If a, b, c and d are in G.P show that $(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (a b + b c + c d)^{2} .$

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Solution

Let 'r' be the common ratio of the given G.P

Then $b = a r, c = a r^{2} a n d d = a r^{3}$

Now, L.H.S $= (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})$

= $(a^{2} + a^{2} r^{2} + a^{2} r^{4}) (a^{2} r^{2} + a^{2} r^{4} + a^{2} r^{6})$

= $a^{2} (1 + r^{2} + r^{4}) a^{2} r^{2} (1 + r^{2} r^{4})$

= $a^{4} r^{2} (1 + r^{2} + r^{4})^{2}$

R.H.S = $(a b + b c + c d)^{2}$

= $(a . a r + a r . a r^{2} + a r^{2} . a r^{3})^{2}$

= $(a^{2} r + a^{2} r^{3} + a^{2} r^{5})^{2}$

$\Rightarrow (a^{2} r)^{2} [1 + r^{2} + r^{4}]^{2} \Rightarrow a^{4} r^{2} [1 + r^{2} + r^{4}]^{2}$

Thus, L.H.S. = R.H.S.

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